Symmetric modules over their endomorphism rings
Let R be an arbitrary ring with identity and M a right R-module with S=EndR(M). In this paper, we study right R-modules M having the property for f,g∈EndR(M) and for m∈M, the condition fgm=0 implies gfm=0. We prove that some results of symmetric rings can be extended to symmetric modules for th...
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Інститут прикладної математики і механіки НАН України
2015
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irk-123456789-1542562019-06-16T01:25:06Z Symmetric modules over their endomorphism rings Ungor, B. Kurtulmaz, Y. Halicioglu, S. Harmanci, A. Let R be an arbitrary ring with identity and M a right R-module with S=EndR(M). In this paper, we study right R-modules M having the property for f,g∈EndR(M) and for m∈M, the condition fgm=0 implies gfm=0. We prove that some results of symmetric rings can be extended to symmetric modules for this general setting. 2015 Article Symmetric modules over their endomorphism rings / B. Ungor, Y. Kurtulmaz, S. Halicioglu, A. Harmanci // Algebra and Discrete Mathematics. — 2015. — Vol. 19, № 2. — С. 283–294. — Бібліогр.: 23 назв. — англ. 1726-3255 2010 MSC:13C99, 16D80. http://dspace.nbuv.gov.ua/handle/123456789/154256 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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English |
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Let R be an arbitrary ring with identity and M a right
R-module with S=EndR(M). In this paper, we study right
R-modules M having the property for f,g∈EndR(M) and
for m∈M, the condition fgm=0 implies gfm=0. We prove
that some results of symmetric rings can be extended to symmetric
modules for this general setting. |
| format |
Article |
| author |
Ungor, B. Kurtulmaz, Y. Halicioglu, S. Harmanci, A. |
| spellingShingle |
Ungor, B. Kurtulmaz, Y. Halicioglu, S. Harmanci, A. Symmetric modules over their endomorphism rings Algebra and Discrete Mathematics |
| author_facet |
Ungor, B. Kurtulmaz, Y. Halicioglu, S. Harmanci, A. |
| author_sort |
Ungor, B. |
| title |
Symmetric modules over their endomorphism rings |
| title_short |
Symmetric modules over their endomorphism rings |
| title_full |
Symmetric modules over their endomorphism rings |
| title_fullStr |
Symmetric modules over their endomorphism rings |
| title_full_unstemmed |
Symmetric modules over their endomorphism rings |
| title_sort |
symmetric modules over their endomorphism rings |
| publisher |
Інститут прикладної математики і механіки НАН України |
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2015 |
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http://dspace.nbuv.gov.ua/handle/123456789/154256 |
| citation_txt |
Symmetric modules over their endomorphism rings / B. Ungor, Y. Kurtulmaz, S. Halicioglu, A. Harmanci // Algebra and Discrete Mathematics. — 2015. — Vol. 19, № 2. — С. 283–294. — Бібліогр.: 23 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
AT ungorb symmetricmodulesovertheirendomorphismrings AT kurtulmazy symmetricmodulesovertheirendomorphismrings AT halicioglus symmetricmodulesovertheirendomorphismrings AT harmancia symmetricmodulesovertheirendomorphismrings |
| first_indexed |
2025-07-14T05:54:49Z |
| last_indexed |
2025-07-14T05:54:49Z |
| _version_ |
1837600594182799360 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 19 (2015). Number 2, pp. 283–294
© Journal “Algebra and Discrete Mathematics”
Symmetric modules
over their endomorphism rings
B. Ungor, Y. Kurtulmaz, S. Halicioglu, A. Harmanci
Communicated by V. Mazorchuk
Abstract. Let R be an arbitrary ring with identity and
M a right R-module with S = EndR(M). In this paper, we study
right R-modules M having the property for f, g ∈ EndR(M) and
for m ∈ M , the condition fgm = 0 implies gfm = 0. We prove
that some results of symmetric rings can be extended to symmetric
modules for this general setting.
1. Introduction
Throughout this paper R denotes an associative ring with identity,
and modules are unitary right R-modules. All right-sided concepts and
results have left-sided counterparts. For a module M , S = EndR(M)
denotes the ring of right R-module endomorphisms of M . Then M is a
left S-module, right R-module and (S, R)-bimodule. In this work, for the
(S, R)-bimodule M , rR(.) and lM (.) denote the right annihilator of a subset
of M in R and the left annihilator of a subset of R in M , respectively.
Similarly, lS(.) and rM (.) are the left annihilator of a subset of M in S
and the right annihilator of a subset of S in M , respectively.
A ring is reduced if it has no nonzero nilpotent elements. In [13],
Krempa introduced the notion of the rigid endomorphism of a ring. An
2010 MSC: 13C99, 16D80.
Key words and phrases: symmetric modules, reduced modules, rigid modules,
semicommutative modules, abelian modules, Rickart modules, principally projective
modules.
284 Symmetric modules over their endomorphism rings
endomorphism α of a ring R is said to be rigid if aα(a) = 0 implies a = 0
for a ∈ R. According to Hong-Kim-Kwak [11], R is said to be an α-rigid
ring if there exists a rigid endomorphism α of R. In [15], a ring R is
symmetric if for any a, b, c ∈ R, abc = 0 implies bac = 0. This is equivalent
to abc = 0 implies acb = 0. A ring R is called semicommutative if for
any a, b ∈ R, ab = 0 implies aRb = 0. A ring R is called abelian if every
idempotent is central, that is, ae = ea for any e2 = e, a ∈ R.
The reduced ring concept was extended to modules by Lee and Zhou
in [16], that is, a right R-module M is called reduced if for any m ∈ M
and any a ∈ R, ma = 0 implies mR ∩ Ma = 0. Similarly, in [2] and
[3], Harmanci et al. extended the rigid ring notion to modules. A right
R-module M is called rigid if for any m ∈ M and any a ∈ R, ma2 = 0
implies ma = 0. Reduced modules are certainly rigid, but the converse is
not true in general. A right R-module M is said to be semicommutative
if for any m ∈ M and any a ∈ R, ma = 0 implies mRa = 0. Abelian
modules are introduced in the context by Roos in [21] and studied by
Goodearl and Boyle in [9]. A module M is called abelian if for any f ∈ S,
e2 = e ∈ S, m ∈ M , we have fem = efm. Note that M is an abelian
module if and only if S is an abelian ring. The concept of (quasi-)Baer rings
was extended by Rizvi and Roman [19] to the general module theoretic
setting, by considering a right R-module M as an (S, R)-bimodule. A
module M is called Baer if for all submodules N of M , lS(N) = Se with
e2 = e ∈ S. A submodule N of M is said to be fully invariant if it is also
a left S-submodule of M . Then the module M is said to be quasi-Baer if
for all fully invariant submodules N of M , lS(N) = Se with e2 = e ∈ S.
Motivated by Rizvi and Roman’s work on (quasi-)Baer modules, the
notion of principally quasi-Baer modules initially appeared in [22]. The
module M is called principally quasi-Baer if for any m ∈ M , lS(Sm) = Sf
for some f2 = f ∈ S. Finally, the concept of right Rickart rings (or right
principally projective rings) was extended to modules in [20], that is,
the module M is called Rickart if for any f ∈ S, rM (f) = eM for some
e2 = e ∈ S, equivalently, Kerf is a direct summand of M .
In this paper, we investigate some properties of symmetric modules
over their endomorphism rings. We prove that if M is a symmetric
module, then S is a symmetric ring. The converse is true for Rickart or
1-epiretractable (in particular, free or regular) or principally projective
modules. Among others it is shown that M is a symmetric module in one of
the cases: (1) S is a strongly regular ring, (2) E(M) is a symmetric module
where E(M) is the injective hull of M . Also, we give a characterization
B. Ungor, Y. Kurtulmaz, S. Halicioglu, A. Harmanci 285
of symmetric rings in terms of symmetric modules, that is, a ring is
symmetric if and only if every cyclic projective module is symmetric.
In what follows, by Z, Q, Zn and Z/nZ we denote, respectively, integers,
rational numbers, the ring of integers modulo n and the Z-module of
integers modulo n.
2. Symmetric modules
Let M be a simple module. By Schur’s Lemma, S = EndR(M) is a
division ring and clearly for any m ∈ M and f , g ∈ S, fgm = 0 implies
gfm = 0. Also every module with a commutative endomorphism ring
satisfies this property. A right R-module M is called R-symmetric ([15]
and [18]) if whenever a, b ∈ R, m ∈ M satisfy mab = 0, we have mba = 0.
R-symmetric modules are also studied by the last two authors of this
paper in [2]. Motivated by this we investigate properties of the class of
modules which are symmetric over their endomorphism rings.
Definition 2.1. Let M be an R-module with S = EndR(M). The module
M is called S-symmetric whenever fgm = 0 implies gfm = 0 for any
m ∈ M and f , g ∈ S.
From now on S-symmetric modules will be called symmetric for the
sake of shortness. Note that a submodule of a symmetric module need
not be symmetric. Therefore we can give the following definition.
Definition 2.2. Let M be an R-module with S = EndR(M) and N an
R-submodule of M . The module N is called a symmetric submodule of
M whenever fgn = 0 implies gfn = 0 for any n ∈ N and f , g ∈ S.
We mention some examples of modules that are symmetric over their
endomorphism rings.
Examples 2.3. (1) Let M be a cyclic torsion Z-module. Then M is
isomorphic to the Z-module (Z/Zpn1
1 ) ⊕ (Z/Zpn2
2 ) ⊕ . . . ⊕ (Z/Zpnt
t ) where
pi (i = 1, . . . , t) are distinct prime integers and ni (i = 1, . . . , t) are positive
integers. EndZ(M) is isomorphic to the commutative ring (Zp
n1
1
)⊕(Zp
n2
2
)⊕
. . . ⊕ (Zp
nt
t
). So M is a symmetric module.
(2) Let p be any prime integer and M = (Z/pZ) ⊕ Q a Z-module.
Then S is isomorphic to the matrix ring
{[
a 0
0 b
]
| a ∈ Zp, b ∈ Q
}
and
so M is a symmetric module.
286 Symmetric modules over their endomorphism rings
There are symmetric modules of which their endomorphism rings are
symmetric, namely simple modules and vector spaces. Our next endeavor
is to find conditions, under which the property of M being symmetric is
equivalent to S being symmetric. A module M is called n-epiretractable [8]
if every n-generated submodule of M is a homomorphic image of M . We
show that Rickart modules and 1-epiretractable modules play an important
role in this direction.
Theorem 2.4. If M is a symmetric module, then S is a symmetric ring.
The converse holds if M satisfies any of the following conditions.
(1) M is a Rickart module.
(2) M is a 1-epiretractable module.
Proof. Let f, g, h ∈ S with fgh = 0. Since M is symmetric, 0 = (fg)hm =
(gf)hm for all m ∈ M . Then gfh = 0. Hence S is symmetric. Conversely,
let M be a Rickart module with fgm = 0 for f, g ∈ S and m ∈ M . Since
M is a Rickart module, there exists e2 = e ∈ S such that rM (fg) = eM .
Hence fge = 0. There exists m′ ∈ M such that m = em′. By multiplying
em′ from the left by e, we have em = eem′ = em′ = m. By using
symmetricity of S repeatedly, it can be easily seen that 0 = fge = 1f(ge)
implies 1(ge)f = gef = 0 and then gfe = 0. Hence gfm = gfem = 0.
Thus M is symmetric. Assume now that M is 1-epiretractable. Then
there exists h ∈ S such that mR = hM . Then we have fghM = 0, and
so fgh = 0. Since S is symmetric, gfh = 0. This implies that gfm = 0.
Therefore M is symmetric.
Corollary 2.5. A free R-module is symmetric if and only if its endo-
morphism ring is symmetric.
Proof. Let F be a free R-module. Clearly, for any m ∈ F there exists
f ∈ EndR(F ) such that fF = mR. Thus F is a 1-epiretractable module.
Therefore Theorem 2.4(2) completes the proof.
Recall that a ring R is said to be regular if for any a ∈ R there exists
b ∈ R with a = aba, while a ring R is called strongly regular if for any
a ∈ R there exists b ∈ R such that a = a2b. It is well known that a ring
is strongly regular if and only if it is reduced and regular (see [14]). Also
every reduced ring is symmetric by [5, Theorem I.3]. Then we have the
following result.
Corollary 2.6. If S is a strongly regular ring, then M is a symmetric
module.
B. Ungor, Y. Kurtulmaz, S. Halicioglu, A. Harmanci 287
Proof. Assume that S is a strongly regular ring. Then S is a symmetric
and regular ring. By [4, Proposition 2.6], M is a Rickart module. The
rest is clear from Theorem 2.4.
A module M is called regular (in the sense of Zelmanowitz [23]) if
for any m ∈ M there exists a right R-homomorphism M
φ
→ R such that
m = mφ(m). Then we have the following result.
Corollary 2.7. If M is a regular module, then the following are equiva-
lent.
(1) M is a symmetric module.
(2) S is a symmetric ring.
Proof. Every cyclic submodule of a regular module is a direct summand,
and so it is 1-epiretractable. It follows from Theorem 2.4.
In [7], Evans introduced principally projective modules as follows: An
R-module M is called principally projective if for any m ∈ M , rR(m) = eR,
where e2 = e ∈ R. The ring R is called right principally projective [10]
if the right R-module R is principally projective. The concept of left
principally projective rings is defined similarly.
In this note, we call the module M principally projective if M is
principally projective as a left S-module, that is, for any m ∈ M , lS(m) =
Se for some e2 = e ∈ S.
It is straightforward that all Baer modules are principally projective.
However quasi-Baer modules need not be principally projective. Namely,
matrix rings over a commutative domain R are quasi-Baer rings; but if
the commutative domain R is not Prüfer, matrix rings over R will not be
principally projective rings. And every quasi-Baer module is principally
quasi-Baer. There are principally projective modules which may not be
quasi-Baer or Baer (see [6, Example 8.2]).
Example 2.8. Let R be a Prüfer domain (a commutative ring with an
identity, no zero divisors, and all finitely generated ideals are projective)
and M denote the right R-module R ⊕ R. By ([12], page 17), S is a 2 × 2
matrix ring over R and it is a Baer ring. Hence M is Baer and so a
principally projective module.
Note that the endomorphism ring of a principally projective module
may not be a right principally projective ring in general. For if M is a
principally projective module and g ∈ S, then we distinguish the two
288 Symmetric modules over their endomorphism rings
cases: Kerg = 0 and Kerg 6= 0. If Kerg = 0, then for any f ∈ rS(g), gf = 0
implies f = 0. Hence rS(g) = 0. Assume that Kerg 6= 0. There exists
a nonzero m ∈ M such that gm = 0. By hypothesis, g ∈ lS(m) = Se
for some e2 = e ∈ S. In this case g = ge and so rS(g) 6 (1 − e)S. The
following example shows that this inclusion is strict.
Example 2.9. Let Q be the ring and N the Q-module constructed by
Osofsky in [17]. Since Q is commutative, we can just as well think of N
as a right Q-module. If S = EndQ(N), then N is a principally projective
module. Identify S with the ring
[
Q 0
Q/I Q/I
]
in the obvious way, and
consider ϕ =
[
0 0
1 + I 0
]
∈ S. Then rS(ϕ) =
[
I 0
Q/I Q/I
]
. This is
not a direct summand of S because I is not a direct summand of Q.
Therefore S is not a right principally projective ring.
Theorem 2.10. If M is a principally projective module, then the following
are equivalent.
(1) M is a symmetric module.
(2) S is a symmetric ring.
Proof. (2) ⇒ (1) Let S be a symmetric ring and assume that fgm = 0 for
some f, g ∈ S and m ∈ M . Since M is principally projective, there exists
e2 = e ∈ S such that lS(gm) = Se. Due to f ∈ lS(gm), we have f = fe
and egm = 0. Similarly, there exists an idempotent e1 ∈ S such that
lS(m) = Se1. Since eg ∈ lS(m), eg = ege1 and e1m = 0. By hypothesis,
Se1m = 0 implies e1Sm = 0 and so ege1Sm = egSm = 0. Note that
symmetric rings are abelian (indeed, since ae(1 − e) = 0 = a(1 − e)e for
any e = e2, a ∈ S, we have ea(1 − e) = 0 = (1 − e)ae. This implies that
ea = ae). Hence 0 = egfm = gfem = gfm. Therefore M is symmetric.
(1) ⇒ (2) Clear.
A proof of the following proposition can be given in the same way as
the proof of [3, Lemma 2.12].
Proposition 2.11. If M is a symmetric module and m ∈ M, fi ∈ S for
1 6 i 6 n, then f1 . . . fnm = 0 if and only if fσ(1) . . . fσ(n)m = 0, where
n ∈ N and σ ∈ Sn.
Lemma 2.12 is a corollary to Lemma 2.18. But we give a proof in
detail.
B. Ungor, Y. Kurtulmaz, S. Halicioglu, A. Harmanci 289
Lemma 2.12. If M is a symmetric module and N a direct summand of
M , then N is a symmetric module.
Proof. Let S1 =EndR(N) and M = N ⊕ K for some submodule K of
M . Let f, g ∈ S1 and n ∈ N with fgn = 0. Define f1(n, k) = (fn, 0)
and g1(n, k) = (gn, 0) where f1, g1 ∈ S = EndR(M), k ∈ K. Then
f1g1(n, 0) = f1(gn, 0) = (fgn, 0) = (0, 0). Since M is symmetric and
f1, g1 ∈ S, g1f1(n, 0) = (0, 0). But (0, 0) = g1f1(n, 0) = g1(fn, 0) =
(gfn, 0). Hence gfn = 0. Therefore N is symmetric.
Corollary 2.13. Let R be a symmetric ring and e ∈ R an idempotent.
Then eR is a symmetric module.
Theorem 2.14. Let R be a ring. Then the following conditions are
equivalent.
(1) Every free R-module is symmetric.
(2) Every projective R-module is symmetric.
Proof. (1) ⇒ (2) Let M be a projective R-module. Then M is a direct
summand of a free R-module F . By (1), F is symmetric and so is M from
Lemma 2.12.
(2) ⇒ (1) Clear.
Theorem 2.15. A ring R is symmetric if and only if every cyclic pro-
jective R-module is symmetric.
Proof. The sufficiency is clear. For the necessity, let M be a cyclic projec-
tive R-module. Then M ∼= I for some direct summand right ideal I of R.
Since R is symmetric, by Lemma 2.12, I is symmetric and so is M .
Any direct sum of symmetric modules need not be symmetric, as the
following example shows.
Example 2.16. Consider the Z-modules Z/2Z and Z/4Z. Clearly, these
modules are symmetric. Let M denote the Z-module Z/2Z⊕Z/4Z. Then
the endomorphism ring EndZ(M) of M is
[
Z2 Z2
Z2 Z4
]
. Consider f =
[
0 1
0 1
]
, g =
[
1 0
0 0
]
and e =
[
1 0
0 1
]
of EndZ(M). Then fge = 0
but gfe 6= 0. Hence EndZ(M) is not a symmetric ring. By Theorem 2.4,
M is not a symmetric module.
290 Symmetric modules over their endomorphism rings
Proposition 2.17. Let M1 and M2 be modules over a ring R. If M1 and
M2 are symmetric and HomR(Mi, Mj) = 0 for i 6= j, then M1 ⊕ M2 is a
symmetric module.
Proof. Let M = M1 ⊕ M2 and Si = EndR(Mi) for i = 1, 2. We may
describe S as
[
S1 0
0 S2
]
. Let f =
[
f1 0
0 f2
]
, g =
[
g1 0
0 g2
]
∈ S with
f1, g1 ∈ S1 and f2, g2 ∈ S2 and m = (m1, m2) ∈ M with m1 ∈ M1, m2 ∈
M2 such that fgm = 0. Then we have f1g1m1 = 0 and f2g2m2 = 0. Since
M1 and M2 are symmetric, g1f1m1 = 0 and g2f2m2 = 0. This implies
that gfm = 0. Therefore M is symmetric.
Lemma 2.18. Let M be an R-module and N a submodule of M . If
M is symmetric and every endomorphism of N can be extended to an
endomorphism of M , then N is also symmetric.
Proof. Let S = EndR(M) and f, g ∈ EndR(N), n ∈ N with fgn = 0. By
hypothesis, there exist α, β ∈ S such that α|N = f and β|N = g. Then
α|N β|N n = 0, and so αβn = 0. Since M is symmetric, we have βαn = 0.
This and αn ∈ N imply that 0 = β|N α|N n = gfn. Therefore N is a
symmetric module.
It is well known that every endomorphism of any module M can
be extended to an endomorphism of the injective hull E(M) of M . By
considering this fact, we can say the next result.
Theorem 2.19. Let M be a module. If E(M) is symmetric, then so is M .
Proof. Clear from Lemma 2.18.
Recall that a module M is quasi-injective if it is M -injective. Then
we have the following.
Theorem 2.20. Let M be a quasi-injective module. If M is symmetric,
then so is every submodule of M .
Proof. Let N be a submodule of M and f ∈ EndR(N). By quasi-injectivity
of M , f extends to an endomorphism of M . Lemma 2.18 completes the
proof.
Let M be an R-module with S = EndR(M). Consider
T (SM) = {m ∈ M | fm = 0 for some nonzero f ∈ S}.
The subset T (SM) of M need not be a submodule of the modules SM
and MR in general, as the following example shows.
B. Ungor, Y. Kurtulmaz, S. Halicioglu, A. Harmanci 291
Example 2.21. Let eij denote 3 × 3 matrix units and consider the ring
R = {(e11 + e22 + e33)a + e12b + e13c + e23d : a, b, c, d ∈ Z2} and the
R-module M = {e12a + e13b + e23c : a, b, c ∈ Z2}. Let f, g ∈ S defined by
f(e12a+e13b+e23c) = e12a+e13b and g(e12a+e13b+e23c) = (e13 +e23)c.
For m = e231, m′ = e121 ∈ M , fm = 0 and gm′ = 0. But no nonzero
elements of S annihilate m + m′ since (m + m′)R = M . Therefore T (SM)
is not a submodule of the modules SM and MR.
In the symmetric case we have the following.
Proposition 2.22. If M is a symmetric module and S is a domain, then
T (SM) is a left S-submodule of M .
Proof. Let m1, m2 ∈ T (SM). There exist nonzero f1, f2 ∈ S with
f1m1 = 0 and f2m2 = 0. Then f1f2m2 = 0. By hypothesis, 0 = f2f1m1 =
f1f2m1. Since S is a domain, f1f2 6= 0 and so f1f2(m1 − m2) = 0 or
m1 − m2 ∈ T (SM). If g ∈ S, then gf1m1 = 0. Since M is symmetric,
gf1m1 = 0 implies f1gm1 = 0. Hence gm1 ∈ T (SM) and so T (SM) is a
left S-submodule of M .
Theorem 2.23. Let M be an R-module with S a domain. Then M is a
symmetric module if and only if T (SM) is a symmetric submodule of M .
Proof. Assume that M is a symmetric module and m ∈ T (SM). There
exists a nonzero f ∈ S with fm = 0. For any r ∈ R, f(mr) = (fm)r = 0.
So mr ∈ T (SM). Therefore T (SM) is an R-submodule of M . Let f, g ∈ S
and m ∈ T (SM) with fgm = 0. Since M is symmetric, gfm = 0 and so
T (SM) is a symmetric submodule of M .
Conversely, let m ∈ M and f, g be nonzero elements of S with fgm = 0.
If m ∈ T (SM), by the symmetry condition on T (SM), we have gfm = 0.
If m 6∈ T (SM), then fg = 0. Since S is a domain, we have a contradiction.
Therefore M is a symmetric module.
Let M be an R-module with S = EndR(M) and N a submodule of
M . The quotient module M/N is called S-symmetric if fgm ∈ N implies
gfm ∈ N for any m ∈ M and f, g ∈ S.
Theorem 2.24. Let M be an R-module with S a domain. If M is sym-
metric, then the quotient module M/T (SM) is S-symmetric.
Proof. Let m ∈ M and f, g ∈ S with fgm ∈ T (SM). So there exists
nonzero h ∈ S such that hfgm = 0. By Proposition 2.11, we have hfgm =
hgfm = 0. Then gfm ∈ T (SM). Hence M/T (SM) is S-symmetric.
292 Symmetric modules over their endomorphism rings
Recall that a module M is called quasi-projective if it is M -projective.
Theorem 2.25. Let M be a module and N a submodule of M .
(1) If M is a quasi-projective module and M/N is S-symmetric, then
M/N is symmetric as a left EndR(M/N)-module.
(2) If N is a fully invariant submodule of M and M/N is symmetric
as a left EndR(M/N)-module, then M/N is S-symmetric.
Proof. (1) Let f1, g1 ∈ EndR(M/N) and m ∈ M with f1g1(m+N) = 0+N
and π denote the natural projection from M to M/N . Since M is quasi-
projective, there exist f, g ∈ S such that f1π = πf and g1π = πg.
Then we have 0 + N = f1g1(m + N) = f1g1πm = f1πgm = πfgm,
and so fgm ∈ N . Hence gfm ∈ N by hypothesis. This implies that
πgfm = g1πfm = g1f1πm = g1f1(m + N) = 0 + N . Therefore M/N is
symmetric as a left EndR(M/N)-module.
(2) Let f, g ∈ S and m ∈ M with fgm ∈ N and π denote the
natural projection from M to M/N . Since N is fully invariant, there
exist f, g ∈ EndR(M/N) such that fπ = πf and gπ = πg. It follows that
fg(m + N) = 0, and so gf(m + N) = 0. Therefore gfm ∈ N .
Proposition 2.26 follows from [1, Theorem 2.14] and [4, Theorem 2.25].
Proposition 2.26. If M is a principally projective module, then the
following conditions are equivalent.
(1) M is a rigid module.
(2) M is a reduced module.
(3) M is a symmetric module.
(4) M is a semicommutative module.
(5) M is an abelian module.
Remark 2.27. It follows from Theorem 2.14 of [1], every reduced module
is semicommutative, and every semicommutative module is abelian. The
converses hold for principally projective modules. Note that for a prime
integer p the cyclic group M of p2 elements is a Z-module for which
S = Zp2 . The module M is neither reduced nor principally projective
although it is semicommutative.
Every symmetric module has a symmetric endomorphism ring. How-
ever, despite all our efforts we have not succeeded in answering positively
the following question for an arbitrary module.
B. Ungor, Y. Kurtulmaz, S. Halicioglu, A. Harmanci 293
Question. Is any module symmetric if its endomorphism ring is symmet-
ric?
The answer is positive for simple modules, vector spaces and the
modules which satisfy the conditions in Theorem 2.4. But if the answer
is negative for an arbitrary module, then what is the counterexample?
Acknowledgements
The authors would like to thank the referee(s) for careful reading of
the manuscript and valuable suggestions. The first author thanks the
Scientific and Technological Research Council of Turkey (TUBITAK) for
the grant.
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Contact information
Burcu Ungor,
Sait Halicioglu
Department of Mathematics,
Ankara University, Turkey
E-Mail(s): bungor@science.ankara.edu.tr,
halici@ankara.edu.tr
Yosum Kurtulmaz Department of Mathematics,
Bilkent University, Turkey
E-Mail(s): yosum@fen.bilkent.edu.tr
Abdullah Harmanci Department of Maths,
Hacettepe University, Turkey
E-Mail(s): harmanci@hacettepe.edu.tr
Received by the editors: 05.01.2013
and in final form 05.12.2014.
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